 | John Radford Young - 1830 - 390 pàgines
...is the same as that for FP • FP in (79), hence as in the ellipse the product of the radii vectores of any point is equal to the square of the semi-diameter conjugate to that passing through the point. PROBLEM II. (87.) The axes of the curve and the inclination of a system... | |
 | Henry Parr Hamilton - 1834 - 272 pàgines
...- b 2 = a.'* - b' 2 , the conjugate diameters cannot be equal to each other in the hyperbola. 239. The rectangle contained by the focal distances of...semi-diameter, conjugate to that which passes through the proposed point. Let P be any point, CD the semi-diameter conjugate to CP, join P, S and P, H ; to prove... | |
 | John Hymers - 1837 - 486 pàgines
...sin 7 = Ci> . PF = 06. 146. Also if we denote CQ or PF by p, we have 'CD i ~ a" + V 1 - a' 2 ' 147- The rectangle contained by the focal distances of any point is equal to the square of the corresponding semiconjugate diameter. CD 2 = a 2 + 6 2 - CP 5 (fig. 40) = a* + b* - л 2 - b 2 + -,... | |
 | Henry Parr Hamilton - 1843 - 314 pàgines
...sin <y is a maximum, that is, when y = — , or when the conjugate diameters are rectangular. 193. The rectangle contained by the focal distances of any point is equal to the square of the corresponding semiconjugate diameter. Let P be any point (at, y), CD the semi-diameter conjugate to... | |
 | Nathan Scholfield - 1845 - 506 pàgines
...Prop. XIV. = CP' + CD' .-. 4AC'+4BC°=4CP'+4CD' Or, A PROPOSITION XVI. THEOREM. The rectangle under the focal distances of any point is equal to the square of the semi-conjugate. That is. if CD be conjugate to CP, SP . HP = CD'. Draw SY. HZ, perpendiculars to the... | |
 | Nathan Scholfield - 1845 - 542 pàgines
...Cor. to Prop. XIV, = CP' + CD' .-. 4AC'+4BC'=4CP'+4CD' PROPOSITION XVI. THEOREM. The rectangle under the focal distances of any point is equal to the square of the semi-conjugate. That is, if CD be conjugate to CP, SP . HP = CD'. Draw SY. HZ, perpendiculars to the... | |
 | Nathan Scholfield - 1845 - 244 pàgines
...+ Cm' + PM' + Dm' = CP + CD' .-. 4AC'+4BC'=4CP'+4CD' PROPOSITION XVI. THEOREM. The rectangle under the focal distances of any point is equal to the square of the semi-conjugate. That is. if CD be conjugate to CP, SP . HP = CD'. Draw SY. HZ, perpendiculars to the... | |
 | Nathan Scholfield - 1845 - 894 pàgines
...Prop. XIV. = CP 4- CD 1 .-. 4AC J +4BC'=4CP"+4CD' Or, A PROPOSITION XVI. THEOREM. The rectangle under the focal distances of any point is equal to the square of the semi-conjugate. That is. if CD be conjugate to CP, Draw SY. HZ, perpendiculars to the tangent at P,... | |
 | Encyclopaedia - 1845 - 404 pàgines
...in the curve, i. "58 ; of hyperbola, rectangle formed by distance* of. from any point in the carve, is equal to the square of the semi-diameter, conjugate to that which pasees through the point, 764 by lenses, 453 by reflection, iii. 432 ; by refraction, 448 ; iv. 379... | |
 | John Radford Young - 1848 - 300 pàgines
...(a/, y ' ) be F'P, FP. then (47) F' P . FP = A 2 — eV 2 . Hence the product of the radii vectores of any point is equal to the square of the semi-diameter conjugate to that passing through the point, that is FT . FP = B '2 . PROBLEM IV. (55). The axes and the inclination... | |
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The rectangle contained by the focal distances of any point, is equal to the square of the semi-diameter conjugate to that which passes through the 
